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Python recursion systematic ordering

I wrote my code and it's working perfectly but the output doesn't really look good. I was it to look more presentable/systematic. How do I do that? This is the kind of result I'm currently getting:
and this is the type of result I want:
This code is basically to find permutations of whatever is inputted.
def permutations(aSet):
if len(aSet) <= 1: return aSet
all_perms = []
first_element = aSet[0:1]
subset = aSet[1:]
partial = permutations(subset)
for permutation in partial:
for index in range(len(aSet)):
new_perm = list(permutation[:index])
new_perm.extend(first_element)
new_perm.extend(permutation[index:])
all_perms.append(new_perm)
return all_perms
I can't figure out what to try.

You can sort the output array with a custom key function. Here keyFunc converts a permutaiton (list of characters) into a single string to perform lexicographic sorting.
from pprint import pprint
# insert your function here
def keyFunc(char_list):
return ''.join(char_list)
chars = list('dog')
permutation = permutations(chars)
permutation.sort(key=keyFunc)
pprint(permutation)
Output:
[['d', 'g', 'o'],
['d', 'o', 'g'],
['g', 'd', 'o'],
['g', 'o', 'd'],
['o', 'd', 'g'],
['o', 'g', 'd']]

Here's a way to order the permutations differently: for each item in the input array, take it out of the array, find all permutations of the remaining subarray, then prepend this item to each permutation of this subarray. This has the effect of placing permutations with similar prefixes together.
from pprint import pprint
def permutations2(chars):
if len(chars) <= 1: return [chars]
all_perms = []
for idx, char in enumerate(chars):
subarr = chars[:idx] + chars[idx+1:]
subperms = permutations2(subarr)
for subperm in subperms:
new_perm = [char] + subperm
all_perms.append(new_perm)
return all_perms
chars = list('dog')
pprint(permutations2(chars))
Result:
[['d', 'o', 'g'],
['d', 'g', 'o'],
['o', 'd', 'g'],
['o', 'g', 'd'],
['g', 'd', 'o'],
['g', 'o', 'd']]

Unique elements of sublists depending on specific value in sublist

I an trying to select unique datasets from a very large quite inconsistent list.
My Dataset RawData consists of string-items of different length.
Some items occure many times, for example: ['a','b','x','15/30']
The key to compare the item is always the last string: for example '15/30'
The goal is: Get a list: UniqueData with items that occure only once. (i want to keep the order)
Dataset:
RawData = [['a','b','x','15/30'],['d','e','f','g','h','20/30'],['w','x','y','z','10/10'],['a','x','c','15/30'],['i','j','k','l','m','n','o','p','20/60'],['x','b','c','15/30']]
My desired solution Dataset:
UniqueData = [['a','b','x','15/30'],['d','e','f','g','h','20/30'],['w','x','y','z','10/10'],['i','j','k','l','m','n','o','p','20/60']]
I tried many possible solutions for instance:
for index, elem in enumerate(RawData): and appending to a new list if.....
for element in list does not work, because the items are not exactly the same.
Can you help me finding a solution to my problem?
Thanks!

The best way to remove duplicates is to add them into a set. Add the last element into a set as to keep track of all the unique values. When the value you want to add is already present in the set unique do nothing if not present add the value to set unique and append the lst to result list here it's new.
Try this.
new=[]
unique=set()
for lst in RawData:
if lst[-1] not in unique:
unique.add(lst[-1])
new.append(lst)
print(new)
#[['a', 'b', 'x', '15/30'],
['d', 'e', 'f', 'g', 'h', '20/30'],
['w', 'x', 'y', 'z', '10/10'],
['i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', '20/60']]

You could set up a new array for unique data and to track the items you have seen so far. Then as you loop through the data if you have not seen the last element in that list before then append it to unique data and add it to the seen list.
RawData = [['a', 'b', 'x', '15/30'], ['d', 'e', 'f', 'g', 'h', '20/30'], ['w', 'x', 'y', 'z', '10/10'],
['a', 'x', 'c', '15/30'], ['i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', '20/60'], ['x', 'b', 'c', '15/30']]
seen = []
UniqueData = []
for data in RawData:
if data[-1] not in seen:
UniqueData.append(data)
seen.append(data[-1])
print(UniqueData)
OUTPUT
[['a', 'b', 'x', '15/30'], ['d', 'e', 'f', 'g', 'h', '20/30'], ['w', 'x', 'y', 'z', '10/10'], ['i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', '20/60']]

RawData = [['a','b','x','15/30'],['d','e','f','g','h','20/30'],['w','x','y','z','10/10'],['a','x','c','15/30'],['i','j','k','l','m','n','o','p','20/60'],['x','b','c','15/30']]
seen = []
seen_indices = []
for _,i in enumerate(RawData):
# _ -> index
# i -> individual lists
if i[-1] not in seen:
seen.append(i[-1])
else:
seen_indices.append(_)
for index in sorted(seen_indices, reverse=True):
del RawData[index]
print (RawData)

Using a set to filter out entries for which the key has already been seen is the most efficient way to go.
Here's a one liner example using a list comprehension with internal side effects:
UniqueData = [rd for seen in [set()] for rd in RawData if not(rd[-1] in seen or seen.add(rd[-1])) ]

How do I alphabetically sort an array of strings without any sort functions? Python

When solving the following problem:
"Assuming you have a random list of strings (for example: a, b, c, d, e, f, g), write a program that will sort the strings in alphabetical order.
You may not use the sort command."
I run into the problem of running strings through the following code, which sometimes gets me duplicated strings in final list
I am fairly new to python and our class just started to look at numpy, and functions in that module, and im not sure of any being used in the code (except any sort function).
import numpy as np
list=[]
list=str(input("Enter list of string(s): "))
list=list.split()
print() # for format space purposes
listPop=list
min=listPop[0]
newFinalList=[]
if(len(list2)!=1):
while(len(listPop)>=1):
for i in range(len(listPop)):
#setting min=first element of list
min=listPop[0]
if(listPop[i]<=min):
min=listPop[i]
print(min)
listPop.pop(i)
newFinalList.append(min)
print(newFinalList)
else:
print("Only one string inputted, so already alphabatized:",list2)
Expected result of ["a","y","z"]
["a","y","z"]
Actual result...
Enter list of string(s): a y z
a
a
a
['a', 'a', 'a']
Enter list of string(s): d e c
d
c
d
d
['c', 'd', 'd']

Selection sort: for each index i of the list, select the smallest item at or after i and swap it into the ith position. Here's an implementation in three lines:
# For each index i...
for i in range(len(list)):
# Find the position of the smallest item after (or including) i.
j = list[i:].index(min(list[i:])) + i
# Swap it into the i-th place (this is a no-op if i == j).
list[i], list[j] = list[j], list[i]
list[i:] is a slice (subset) of list starting at the ith element.
min(list) gives you the smallest element in list.
list.index(element) gives you the (first) index of element in list.
a, b = b, a atomically swaps the values of a and b.
The trickiest part of this implementation is that when you're using index to find the index of the smallest element, you need to find the index within the same list[i:] slice that you found the element in, otherwise you might select a duplicate element in an earlier part of the list. Since you're finding the index relative to list[i:], you then need to add i back to it to get the index within the entire list.

You can implement Quick sort for same:
def partition(arr,low,high):
i = ( low-1 )
pivot = arr[high]
for j in range(low , high):
if arr[j] <= pivot:
i = i+1
arr[i],arr[j] = arr[j],arr[i]
arr[i+1],arr[high] = arr[high],arr[i+1]
return ( i+1 )
def quickSort(arr,low,high):
if low < high:
pi = partition(arr,low,high)
quickSort(arr, low, pi-1)
quickSort(arr, pi+1, high)
arr = ['a', 'x', 'p', 'o', 'm', 'w']
n = len(arr)
quickSort(arr,0,n-1)
print ("Sorted list is:")
for i in range(n):
print ("%s" %arr[i]),
output:
Sorted array is:
a m o p w x

Mergesort:
from heapq import merge
from itertools import islice
def _ms(a, n):
return islice(a,n) if n<2 else merge(_ms(a,n//2),_ms(a,n-n//2))
def mergesort(a):
return type(a)(_ms(iter(a),len(a)))
# example
import string
import random
L = list(string.ascii_lowercase)
random.shuffle(L)
print(L)
print(mergesort(L))
Sample run:
['h', 'g', 's', 'l', 'a', 'f', 'b', 'z', 'x', 'c', 'r', 'j', 'q', 'p', 'm', 'd', 'k', 'w', 'u', 'v', 'y', 'o', 'i', 'n', 't', 'e']
['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']

Solving a "colored Quxes" coding challenge with recursion

I am trying to solve some of the coding challenges that I find online. However I was stopped by the below problem. I tried to solve it using recursion but I feel I am missing a very important concept in recursion. My code works for all of the below examples except the last one it will break down.
Can someone point to me the mistake that I made in this recursion code? Or maybe guide me through solving the issue?
I know why my code breaks but I don't know how to get around the "pass by object reference" in Python which I think creating the bigger problem for me.
The coding question is:
On a mysterious island there are creatures known as Quxes which come in three colors: red, green, and blue. One power of the Qux is that if two of them are standing next to each other, they can transform into a single creature of the third color.
Given N Quxes standing in a line, determine the smallest number of them remaining after any possible sequence of such transformations.
For example, given the input ['R', 'G', 'B', 'G', 'B'], it is possible to end up with a single Qux through the following steps:
Arrangement | Change
----------------------------------------
['R', 'G', 'B', 'G', 'B'] | (R, G) -> B
['B', 'B', 'G', 'B'] | (B, G) -> R
['B', 'R', 'B'] | (R, B) -> G
['B', 'G'] | (B, G) -> R
['R'] |
________________________________________
My code is:
class fusionCreatures(object):
"""Regular Numbers Gen.
"""
def __init__(self , value=[]):
self.value = value
self.ans = len(self.value)
def fusion(self, fus_arr, i):
color = ['R','G','B']
color.remove(fus_arr[i])
color.remove(fus_arr[i+1])
fus_arr.pop(i)
fus_arr.pop(i)
fus_arr.insert(i, color[0])
return fus_arr
def fusionCreatures1(self, arr=None):
# this method is to find the smallest number of creature in a row after fusion
if arr == None:
arr = self.value
for i in range (0,len(arr)-1):
#print(arr)
if len(arr) == 2 and i >= 1 or len(arr)<2:
break
if arr[i] != arr[i+ 1]:
arr1 = self.fusion(arr, i)
testlen = self.fusionCreatures1(arr)
if len(arr) < self.ans:
self.ans = len(arr)
return self.ans
Testing array (all of them work except the last one):
t1 = fusionCreatures(['R','G','B','G','B'])
t2 = fusionCreatures(['R','G','B','R','G','B'])
t3 = fusionCreatures(['R','R','G','B','G','B'])
t4 = fusionCreatures(['G','R','B','R','G'])
t5 = fusionCreatures(['G','R','B','R','G','R','G'])
t6 = fusionCreatures(['R','R','R','R','R'])
t7 = fusionCreatures(['R', 'R', 'R', 'G', 'G', 'G', 'B', 'B', 'B'])
print(t1.fusionCreatures1())
print(t2.fusionCreatures1())
print(t3.fusionCreatures1())
print(t4.fusionCreatures1())
print(t5.fusionCreatures1())
print(t6.fusionCreatures1())
print(t7.fusionCreatures1())

I'll start by mentioning that there is a deductive approach that works in O(n) and is detailed in this blog post. It boils down to checking the parity of the counts of the three types of elements in the list to determine which of a few fixed outcomes occurs.
You mention that you'd prefer to use a recursive approach, which is O(n!). This is a good start because it can be used as a tool for helping arrive at the O(n) solution and is a common recursive pattern to be familiar with.
Because we can't know whether a given fusion between two Quxes will ultimately lead to an optimal global solution we're forced to try every possibility. We do this by walking over the list and looking for potential fusions. When we find one, perform the transformation in a new list and call fuse_quxes on it. Along the way, we keep track of the smallest length achieved.
Here's one approach:
def fuse_quxes(quxes, choices="RGB"):
fusion = {x[:-1]: [x[-1]] for x in permutations(choices)}
def walk(quxes):
best = len(quxes)
for i in range(1, len(quxes)):
if quxes[i-1] != quxes[i]:
sub = quxes[:i-1] + fusion[quxes[i-1], quxes[i]] + quxes[i+1:]
best = min(walk(sub), best)
return best
return walk(quxes)
This is pretty much the direction your provided code is moving towards, but the implementation seems unclear. Unfortunately, I don't see any single or quick fix. Here are a few general issues:
Putting the fusionCreatures1 function into a class allows it to mutate external state, namely self.value and self.ans. self.value in particular is poorly named and difficult to keep track of. It seems like the intent is to use it as a reference copy to reset arr to its default value, but arr = self.value means that when fus_arr is mutated in fusion(), self.value is as well. Everything is pretty much a reference to one underlying list.
Adding slices to these copies at least makes the program easier to reason about, for example, arr = self.value[:] and fus_arr = fus_arr[:] in the fusion() function. In short, try to write pure functions.
self.ans is also unclear and unnecessary; better to keep the result value relegated to a local variable within the recursive function.
It seems unnecessary to put a stateless function into a class unless it's a purely static method and the class is acting as a namespace.
Another cause of cognitive overload are branching statements like if and break. We want to minimize the frequency and nesting of these. Here is fusionCreatures1 in pseudocode, with annotations for mutations and complex interactions:
def fusionCreatures1():
if ...
read mutated global state
for i in len(arr):
if complex length and index checks:
break
if arr[i] != arr[i+ 1]:
impure_func_that_changes_arr_length(arr)
recurse()
if new best compared to global state:
mutate global state
You'll probably agree that it's pretty difficult to mentally step through a run of this function.
In fusionCreatures1(), two variables are unused:
arr1 = self.fusion(arr, i)
testlen = self.fusionCreatures1(arr)
The assignment arr1 = self.fusion(arr, i) (along with the return fus_arr) seems to indicate a lack of understanding that self.fusion is really an in-place function that mutates its argument array. So calling it means arr1 is arr and we have another aliased variable to reason about.
Beyond this, neither arr1 or testlen are used in the program, so the intent is unclear.
A good linter will pick up these unused variables and identify most of the other complexity issues I've mentioned.
Mutating a list while looping over it is usually disastrous. self.fusion(arr, i) mutates arr inside a loop, making it very difficult to reason about its length and causing an index error when the range(len(arr)) no longer matches the actual len(arr) in the function body (or at least necessitating an in-body precondition). Making self.fusion(arr, i) pure using a slice, as mentioned above, fixes this problem but reveals that there is no recursive base case, resulting in a stack overflow error.
Avoid variable names like arr, arr1, value unless the context is obvious. Again, these obfuscate intent and make the program difficult to understand.
Some minor style suggestions:
Use snake_case per PEP-8. Class names should be TitleCased to differentiate them from functions. No need to inherit from object--that's implicit.
Use consistent spacing around functions and operators: range (0,len(arr)-1): is clearer as range(len(arr) - 1):, for example. Use vertical whitespace around blocks.
Use lists instead of typing out t1, t2, ... t7.
Function names should be verbs, not nouns. A class like fusionCreatures with a method called fusionCreatures1 is unclear. Something like QuxesSolver.minimize(creatures) makes the intent a bit more obvious.
As for the solution I provided above, there are other tricks worth considering to speed it up. One is memoization, which can help avoid duplicate work (any given list will always produce the same minimized length, so we just store this computation in a dict and spit it back out if we ever see it again). If we hit a length of 1, that's the best we can do globally, so we can skip the rest of the search.
Here's a full runner, including the linear solution translated to Python (again, defer to the blog post to read about how it works):
from collections import defaultdict
from itertools import permutations
from random import choice, randint
def fuse_quxes_linear(quxes, choices="RGB"):
counts = defaultdict(int)
for e in quxes:
counts[e] += 1
if not quxes or any(x == len(quxes) for x in counts.values()):
return len(quxes)
elif len(set(counts[x] % 2 for x in choices)) == 1:
return 2
return 1
def fuse_quxes(quxes, choices="RGB"):
fusion = {x[:-1]: [x[-1]] for x in permutations(choices)}
def walk(quxes):
best = len(quxes)
for i in range(1, len(quxes)):
if quxes[i-1] != quxes[i]:
sub = quxes[:i-1] + fusion[quxes[i-1], quxes[i]] + quxes[i+1:]
best = min(walk(sub), best)
return best
return walk(quxes)
if __name__ == "__main__":
tests = [
['R','G','B','G','B'],
['R','G','B','R','G','B'],
['R','R','G','B','G','B'],
['G','R','B','R','G'],
['G','R','B','R','G','R','G'],
['R','R','R','R','R'],
['R', 'R', 'R', 'G', 'G', 'G', 'B', 'B', 'B']
]
for test in tests:
print(test, "=>", fuse_quxes(test))
assert fuse_quxes_linear(test) == fuse_quxes(test)
for i in range(100):
test = [choice("RGB") for x in range(randint(0, 10))]
assert fuse_quxes_linear(test) == fuse_quxes(test)
Output:
['R', 'G', 'B', 'G', 'B'] => 1
['R', 'G', 'B', 'R', 'G', 'B'] => 2
['R', 'R', 'G', 'B', 'G', 'B'] => 2
['G', 'R', 'B', 'R', 'G'] => 1
['G', 'R', 'B', 'R', 'G', 'R', 'G'] => 2
['R', 'R', 'R', 'R', 'R'] => 5
['R', 'R', 'R', 'G', 'G', 'G', 'B', 'B', 'B'] => 2

Here is my suggestion.
First, instead of "R", "G" and "B" I use integer values 0, 1, and 2. This allows nice and easy fusion between a and b, as long as they are different, by simply doing 3 - a - b.
Then my recursion code is:
def fuse_quxes(l):
n = len(l)
for i in range(n - 1):
if l[i] == l[i + 1]:
continue
else:
newn = fuse_quxes(l[:i] + [3 - l[i] - l[i + 1]] + l[i+2:])
if newn < n:
n = newn
return n
Run this with
IN[5]: fuse_quxes([0, 0, 0, 1, 1, 1, 2, 2, 2])
Out[5]: 2

Here is my attempt of the problem
please find the description in comment
inputs = [['R','G','B','G','B'],
['R','G','B','R','G','B'],
['R','R','G','B','G','B'],
['G','R','B','R','G'],
['G','R','B','R','G','R','G'],
['R','R','R','R','R'],
['R', 'R', 'R', 'G', 'G', 'G', 'B', 'B', 'B'],]
def fuse_quxes(inp):
RGB_set = {"R", "G", "B"}
merge_index = -1
## pair qux with next in line and loop through all pairs
for i, (q1, q2) in enumerate(zip(inp[:-1], inp[1:])):
merged = RGB_set-{q1,q2}
## If more than item remained in merged after removing q1 and q2 qux can't fuse
if(len(merged))==1:
merged = merged.pop()
merge_index=i
merged_color = merged
## loop through the pair until result of fuse is different from qux in either right
## or left side
if (i>0 and merged!=inp[i-1]) or ((i+2)<len(inp) and merged!=inp[i+2]):
break
print(inp)
## merge two qux which results to qux differnt from either its right or left else do any
## possible merge
if merge_index>=0:
del inp[merge_index]
inp[merge_index] = merged_color
return fuse_quxes(inp)
else:
## if merge can't be made break the recurssion
print("Result", len(inp))
print("_______________________")
return len(inp)
[fuse_quxes(inp) for inp in inputs]
output
['R', 'G', 'B', 'G', 'B']
['R', 'R', 'G', 'B']
['R', 'B', 'B']
['G', 'B']
['R']
Result 1
_______________________
['R', 'G', 'B', 'R', 'G', 'B']
['R', 'G', 'B', 'R', 'R']
['R', 'G', 'G', 'R']
['B', 'G', 'R']
['B', 'B']
Result 2
_______________________
['R', 'R', 'G', 'B', 'G', 'B']
['R', 'B', 'B', 'G', 'B']
['G', 'B', 'G', 'B']
['R', 'G', 'B']
['R', 'R']
Result 2
_______________________
['G', 'R', 'B', 'R', 'G']
['G', 'G', 'R', 'G']
['G', 'B', 'G']
['R', 'G']
['B']
Result 1
_______________________
['G', 'R', 'B', 'R', 'G', 'R', 'G']
['G', 'G', 'R', 'G', 'R', 'G']
['G', 'B', 'G', 'R', 'G']
['R', 'G', 'R', 'G']
['B', 'R', 'G']
['B', 'B']
Result 2
_______________________
['R', 'R', 'R', 'R', 'R']
Result 5
_______________________
['R', 'R', 'R', 'G', 'G', 'G', 'B', 'B', 'B']
['R', 'R', 'B', 'G', 'G', 'B', 'B', 'B']
['R', 'G', 'G', 'G', 'B', 'B', 'B']
['B', 'G', 'G', 'B', 'B', 'B']
['R', 'G', 'B', 'B', 'B']
['R', 'R', 'B', 'B']
['R', 'G', 'B']
['R', 'R']
Result 2
_______________________
[1, 2, 2, 1, 2, 5, 2]

Change the way sorted works in Python (different than alphanumeric)

I'm representing cards in poker as letters (lower and uppercase) in order to store them efficiently. I basically now need a custom sorting function to allow calculations with them.
What is the fastest way to sort letters in Python using
['a', 'n', 'A', 'N', 'b', 'o', ....., 'Z']
as the ranks rather than
['A', 'B', 'C', 'D', 'E', 'F', ....., 'z']
which is the default?
Note, this sorting is derived from:
import string
c = string.letters[:13]
d = string.letters[13:26]
h = string.letters[26:39]
s = string.letters[39:]
'a' = 2 of clubs
'n' = 2 of diamonds
'A' = 2 of hearts
'N' = 2 of spades
etc

You can provide a key function to sorted, this function will be called for each element in the iterable and the return value will be used for the sorting instead of the elements value.
In this case it might look something like the following:
order = ['a', 'n', 'A', 'N', 'b', 'o', ....., 'Z']
sorted_list = sorted(some_list, key=order.index)
Here is a brief example to illustrate this:
>>> order = ['a', 'n', 'A', 'N']
>>> sorted(['A', 'n', 'N', 'a'], key=order.index)
['a', 'n', 'A', 'N']
Note that to make this more efficient you may want to use a dictionary lookup for your key function instead of order.index, for example:
order = ['a', 'n', 'A', 'N', 'b', 'o', ....., 'Z']
order_dict = {x: i for i, x in enumerate(order)}
sorted_list = sorted(some_list, key=order_dict.get)

Store [edit: not store, use internally] them as numbers ordered by value and only convert to letters when displaying them.
Edit: if 1 byte values are required then you can have the cards in the range 1:52 as characters, then, again, convert to the proper letters when displaying and storing them.